solutions of exponential functional equation

Question

Let $G$ be a group and $m_1, m_2: G\to \Bbb C$ are unbounded functions satisfying the exponential functional equation $$ m_j(x+y)=m_j(x)m_j(y),\,\, j=1, 2 $$ for all $x, y\in G$. If $|m_1(x)-m_2(x)|$ is a bounded function, then is it true that $m_1=m_2$ ?

Answer 1

Since you use "+" for the group operation. I assume we are dealing with an abelian group.

In this case, $m_1$ does coincide with $m_2$.

It is easy to see:

• $\forall x \in G, m_j(x) \ne 0$. $$\begin{align} m_j \text{ unbounded} \implies & \exists y \in G \text{ s.t. }m_j(y) \ne 0\\ \implies & \forall x \in G,\; m_j(x) m_j(y-x) = m_j(y) \ne 0\\ \implies & \forall x \in G,\; m_j(x) \ne 0 \end{align}$$

• $m_j(0) \ne 0 \wedge m_j(0) = m_j(0+0) = m_j(0)^2 \implies m_j(0) = 1$.

• $\forall x \in G$, $m_j(x)m_j(-x) = m_j(x-x) = m_j(0) = 1 \implies m_j(-x) = m_j(x)^{-1}$.

Start with these observations, one can use induction to show:

$$\forall n \in \mathbb{Z}, x \in G,\; m_j(nx) = m_j(x)^n$$

Let $K = \sup \{ | m_1(x) - m_2(x)| : x \in G \}$. For any $x \in G$, there are 3 possible cases:

Case 1 At least one of $|m_j(x)| > 1$.

WOLOG, we will assume $|m_1(x)| \ge |m_2(x)|$ and $> 1$. For any positive integer $n$, we have:

$$|m_1(x)^n - m_2(x)^n | = |m_1(nx) - m_2(nx)| \le K$$ $$\implies \left| 1 - \left(\frac{m_2(x)}{m_1(x)}\right)^n \right| \le \frac{K}{|m_1(x)|^n}\tag{*1}$$

Since R.H.S of $(*1)$ converges to $0$ as $n \to \infty$. This is possible only if $m_1(x) = m_2(x)$.

Case 2 At least one of $|m_j(x)| < 1$.

WOLOG, we will assume $|m_1(x)| \le |m_2(x)|$ and $< 1$. Since $m_j(-x) = m_j(x)^{-1}$, we see $m_1(-x)$ and $m_2(-x)$ satisfy condition in Case 1. This means:

$$m_1(-x) = m_2(-x) \implies m_1(x) = m_1(-x)^{-1} = m_2(-x)^{-1} = m_2(x)$$

Case 3 $|m_1(x)| = |m_2(x)| = 1$.

Since $m_1$ is unbounded, pick a $y \in G$ such that $|m_1(y)| > 1$, we have:

$$| m_1(y+x) | = |m_1(y)||m_1(x)| = |m_1(y)| >1$$

This means $m_1(y+x)$ and $m_2(y+x)$ satisfy condition in Case 1. Together with the obvious equality $m_1(y) = m_2(y)$, we get: $$m_1(y+x) = m_2(y+x) \implies m_1(x) = \frac{m_1(y+x)}{m_1(y)} = \frac{m_2(y+x)}{m_2(y)} = m_2(x)$$